Minimum sum of two numbers formed from digits of an array in O(n)
Last Updated :
20 Feb, 2023
Given an array of digits (values are from 0 to 9), find the minimum possible sum of two numbers formed from digits of the array. All digits of the given array must be used to form the two numbers.
Examples:
Input: arr[] = {6, 8, 4, 5, 2, 3}
Output: 604
246 + 358 = 604
Input: arr[] = {5, 3, 0, 7, 4}
Output: 82
Approach: A minimum number will be formed from the set of digits when smallest digit appears at the most significant position and next to smallest digit appears at next most significant position and so on…
The idea is to build two numbers by alternating picking digits from the array (assuming it is sorted in ascending). So the first number is formed by digits present in odd positions in the array and the second number is formed by digits from even positions in the array. Finally, we return the sum of the first and second number. In order to reduce the time complexity, the array can be sorted in O(n) using the frequency array of digits as every element of the original array is a single digit i.e. there can be at most 10 distinct elements.
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>
using namespace std;
int minSum(vector< int > arr, int n)
{
int MAX = 10;
int *freq = new int [MAX];
for ( int i = 0; i < n; i++) {
freq[arr[i]]++;
}
int k = 0;
for ( int i = 0; i < MAX; i++) {
for ( int j = 0; j < freq[i]; j++) {
arr[k++] = i;
}
}
int num1 = 0;
int num2 = 0;
for ( int i = 0; i < n; i++) {
if (i % 2 == 0)
num1 = num1 * MAX + arr[i];
else
num2 = num2 * MAX + arr[i];
}
return num1 + num2;
}
int main( void )
{
vector< int >arr = { 6, 8, 4, 5, 2, 3 };
int n = arr.size();
cout << minSum(arr, n);
}
|
Java
public class GFG {
public static final int MAX = 10 ;
static int minSum( int arr[], int n)
{
int freq[] = new int [MAX];
for ( int i = 0 ; i < n; i++) {
freq[arr[i]]++;
}
int k = 0 ;
for ( int i = 0 ; i < MAX; i++) {
for ( int j = 0 ; j < freq[i]; j++) {
arr[k++] = i;
}
}
int num1 = 0 ;
int num2 = 0 ;
for ( int i = 0 ; i < n; i++) {
if (i % 2 == 0 )
num1 = num1 * MAX + arr[i];
else
num2 = num2 * MAX + arr[i];
}
return num1 + num2;
}
public static void main(String[] args)
{
int arr[] = { 6 , 8 , 4 , 5 , 2 , 3 };
int n = arr.length;
System.out.print(minSum(arr, n));
}
}
|
Python3
def minSum(arr, n):
MAX = 10
freq = [ 0 ] * MAX
for i in range (n):
freq[arr[i]] + = 1
k = 0
for i in range ( MAX ):
for j in range ( 0 ,freq[i]):
arr[k] = i
k + = 1
num1 = 0
num2 = 0
for i in range (n):
if i % 2 = = 0 :
num1 = num1 * MAX + arr[i]
else :
num2 = num2 * MAX + arr[i]
return num1 + num2
arr = [ 6 , 8 , 4 , 5 , 2 , 3 ]
n = len (arr);
print (minSum(arr, n))
|
C#
using System;
class GFG {
public static int MAX = 10;
static int minSum( int [] arr, int n)
{
int [] freq = new int [MAX];
for ( int i = 0; i < n; i++) {
freq[arr[i]]++;
}
int k = 0;
for ( int i = 0; i < MAX; i++) {
for ( int j = 0; j < freq[i]; j++) {
arr[k++] = i;
}
}
int num1 = 0;
int num2 = 0;
for ( int i = 0; i < n; i++) {
if (i % 2 == 0)
num1 = num1 * MAX + arr[i];
else
num2 = num2 * MAX + arr[i];
}
return num1 + num2;
}
static public void Main()
{
int [] arr = { 6, 8, 4, 5, 2, 3 };
int n = arr.Length;
Console.WriteLine(minSum(arr, n));
}
}
|
Javascript
<script>
let MAX = 10;
function minSum(arr, n)
{
let freq = new Array(MAX);
freq.fill(0);
for (let i = 0; i < n; i++) {
freq[arr[i]]++;
}
let k = 0;
for (let i = 0; i < MAX; i++) {
for (let j = 0; j < freq[i]; j++) {
arr[k++] = i;
}
}
let num1 = 0;
let num2 = 0;
for (let i = 0; i < n; i++) {
if (i % 2 == 0)
num1 = num1 * MAX + arr[i];
else
num2 = num2 * MAX + arr[i];
}
return num1 + num2;
}
let arr = [ 6, 8, 4, 5, 2, 3 ];
let n = arr.length;
document.write(minSum(arr, n));
</script>
|
Time Complexity: O(n)
Space Complexity: O(n)
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